An original sample of K-40 has a mass of 25.00 grams.
After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged.
What is the half-life of K-40?
First step is to determine the remaining decimal amount.
3.125 grams /25.00 grams = 0.125
Second step is to determine the number of half lives.
(1/2)^n = 0.125
N log (1/2) = log 0.125
N = 3 years
Answer:
Kp is 0.228/atm
Explanation:
This is the reaction:
CO + Cl<u>₂</u> → CCl₂O
1 mol of carbon monoxide and 1 mol of chlorine produce 1 mol of phosgene.
Formula for Kp which derivates from Kc is:
Kp = Kc (R.T)ⁿᵇ ⁻ ⁿᵃ
Δп = nb (moles in the products) - nₐ(moles in the reactants)
Δп = 1 - 2 = -1
T is T° in K → T°C + 273 = 611°C +273 = 884K
R → Universal constant gas → 0.082 L.atm/mol.K
We replace the data: Kp = 16.5 L/mol (0.082 . 884K)⁻¹ → 0.228/atm
Answer:
I think the answer is 2..
Central Vacuole in plant cells:
- holds materials and wastes.
- provide structure and support for the growing plant.
Hope this helped and plz mark as brainliest!
Answer:
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